That was one of the questions in which I was "retarded" because even though I kind of knew what Einstein's equations were exactly saying in the middle of the high school, it was unclear to me how can someone ever manage to cross the horizon. Fortunately, CommunistSocialistSwine, pretending to be CapitalistImperialistPig in order to damage the capitalists' image, has repeated the same question.

The (wrong) argument that nothing ever manages to get inside the black hole goes as follows. The observer at infinity, Paolo, never sees the infalling observer - that will be referred to as a beer can to preserve the standard conventions here - crossing the horizon because the final moments of the can's live before the horizon crossing are infinitely stretched.

That's fine but even the observer who lives closer to the event horizon, Julio, seems to see the can's moments right before the horizon crossing stretched to infinite time intervals. At some moment, this expansion gets really extreme. Microseconds of the can's life can get stretched to times that are longer than the lifetime of the black hole (that evaporates by the Hawking radiation).

Moreover, as you choose Julio's base to be ever closer to the horizon, he seems to see the black hole evaporate ever faster, as measured by his proper time. You may put him so close to the horizon that it only takes a microsecond of his proper time to see the whole black hole evaporation process.

It's such a short time that you must conclude that the beer can, which is really close to Julio, simply can't have enough time to cross the horizon. Before the beer can manage to cross the horizon, the black hole must surely evaporate, so the beer can (or parts of the stars that is supposed to collapse into a black hole) can never get inside. Or can it?

Because we will need the diagram in the rest of the text, I decided to borrow a technology from the postings with song lyrics and confine the whole remaining text into a scrolling box.

When you want to use the Penrose diagram to settle these questions, you need to understand or check three types of facts:

- The meaning of the Penrose diagrams in general

- The reasons why the particular picture above describes an evaporating black hole arising from a collapsing star
- The tricks to derive all the desired answers from the picture above

A Penrose diagram is a two-dimensional drawing similar to the picture above whose goal is to capture the geometry of a two-dimensional spacetime with Minkowski, special relativistic signature - especially its causal structure, a point we will explain in more detail.

It can also be used to clarify the same information about a higher-dimensional spacetime geometry as long as the remaining dimensions are trivially attached to each point of the two-dimensional diagram.

For example, each point on the picture above is a two-dimensional sphere of some radius (that depends on the point) and the full formula for the geometry, "ds^2", is simply the sum of the "ds^2" from the two-sphere and "ds^2" from the two-dimensional diagram. This simplification of a higher-dimensional geometry into a two-dimensional picture is possible because of the rotational symmetry.

The trick of the diagram is that it must correctly encode all the (hyperbolic, Minkowskian) "angles" in the two-dimensional spacetime, but it is not expected that the actual distances agree: they can be rescaled by a factor that can depend on the point in the diagram.

The angles must be chosen in such a way that all the diagonal lines slanted by 45 degrees must be light-like i.e. null: "ds^2" must be equal to zero, just like you expect in normal pictures of the flat Minkowski space. Correspondingly, all lines that are mostly horizontal are spacelike and all lines that are mostly vertical are time-like i.e. acceptable world lines for massive observers (who can never move faster than light).

Why Penrose diagrams always exist

Now, it is not hard to see that it is always possible to draw a similar picture for a region of a two-dimensional spacetime. Why? Because the two-dimensional geometry is described by the metric tensor, i.e. three independent functions g_{00}, g_{01}, and g_{11} of the coordinates "x^0, x^1".

On the other hand, the "ds^2" geometry is encoded by the Penrose diagram if the diagram is supplemented with one function of the two diagram coordinates, namely with "scale(x,t)" where "x,t" are the horizontal and vertical axes on our Penrose diagram.

So the question is whether there exists a coordinate transformation from the old general coordinates "x^0, x^1" (where the metric tensor was given by its three components) to the special coordinates "x,t" of the Penrose diagram (where the geometry is given by "ds^2 = scale(x,t) (t^2 - x^2)"). Well, obviously it exists, at least locally (in a region) for sensible enough geometries. Why?

Because the task is really to find three functions, "x^0(x,t), x^1(x,t), scale(x,t)" such that the induced geometry converted to the "x^0, x^1" coordinates will coincide with the geometry encoded in "g_{mn} (x^0, x^1)". That's a set of three differential equations for three functions and you may be pretty certain that a solution exists, at least given some reasonable conditions.

This explanation also makes it clear that such an angle-preserving diagrams can't generally exist in three or more spacetime dimensions (if the geometry non-trivially depends on all of them) because the metric tensor will have 6 independent components (in 3 dimensions) which is more than the 4 functions that would determine the geometry of the diagram (one scale plus three new coordinates): in higher dimensions, the counting would get even worse.

By the way, the explanation in the previous paragraph is also the reason why strings from string theory - with their two-dimensional world sheets - are special and, unlike their higher-dimensional counterparts, they allow us to get rid of all the dangerous physical features of the gravitational theory inside the world sheet (unlike the higher-dimensional world volumes).

Uniqueness

Fine. Is the solution - the diagram for a given spacetime geometry - unique? The answer is No. But the non-uniqueness is easy to understand. Consider the two-dimensional spacetime encoded in a Penrose diagram with coordinates "x,t" on the paper. Introduce the coordinates "xPLUS, xMINUS" which are equal to "x+t" and "x-t", respectively. If you remember the formula for the difference of squares, the geometry may be easily written as

ds^2 = scale(xPLUS,xMINUS) xPLUS xMINUSWhat you are allowed to do is to redefine xPLUS to its increasing function yPLUS and xMINUS to its function yMINUS. The metric will continue to factorize simply because "scale" will only be replaced by "dxPLUS/dyPLUS times dxMINUS/dyMINUS", the "Jacobian", and it will continue to be

ds^2 = scale(yPLUS,yMINUS) yPLUS yMINUSThat's also known to be the only freedom you have to redesign a Penrose diagram. For the simplest and most important case of the flat space, xPLUS and xMINUS were real numbers between -infinity and +infinity (much like "x,t" themselves). But you may replace them by their functions yPLUS, yMINUS that go e.g. from -pi/2 to +pi/2, by defining "y" to be "arctan(x)" for both xPLUS and xMINUS.

The function "tanh" would also work (putting the result between -1 and +1) but there are reasons why "arctan" is the popular convention.

That means that the right, finite (or "compactified") Penrose diagram for the Minkowski space is a square rotated by 45 degrees - which is easier to manipulate with than an infinite piece of paper. It's important that the boundaries corresponding to some regions at infinity are light-like - the boundaries of the square.

If you are interested in the four-dimensional (or higher-dimensional) flat spacetime, each point on the "x,t" Penrose diagram corresponds to a sphere (more concretely, a two-dimensional sphere in the ordinary case of four spacetime dimensions).

A spatial coordinate - the old "x" in the Penrose diagram that was infinite, before we "compactified" it - also determines the radius of the sphere hiding at each point. It's the radial coordinate that should never be negative. So the Penrose diagram for the four-dimensional Minkowski space is just one half of the rotated square, i.e. a triangle with a vertical line on the left side. That's where the internal two-spheres shrink to zero.

**Penrose diagram of the Schwarzschild black hole**

Parts of the picture above look like the triangle but there's more structure in it. The bottom of the picture, the "infinite past", looks just like in the flat space. You can see that we also had some star with a surface (the yellow world line). At the beginning, it was larger than the Schwarzschild radius (the orange line, the points where the radius of the internal sphere is "R=2GM/c^2").

The whole "infinite past" of the star is concentrated in the very bottom portions of the yellow line, because of our "arctan" transformation of the light-like coordinates explained above.

The lower part of the picture also contains the diagonal light-like line called the "infinite past" on the picture or "scri minus" (use a script letter "I" with a minus index, and read "scri" normally as "s-cry"). Analogously, the upper, future portion of the picture contains the "infinite future", the "scri plus".

Note that the extreme upper tip of the picture also looks just like in the flat space even though it is moved to the right side relatively to the lower portion of the diagram. That's where the black hole has already evaporated. If you want to consider a black hole that never evaporates, you must shrink the small "tooth" at the top to zero.

If you care, the radius of the hidden sphere shrinks to zero at both vertical lines in the picture and at the horizontal singularity, while it goes mostly to infinity at all the diagonal lines. The only places on the boundary of the diagram where the radius is finite are the upper and lower corner (point!): that's where all the "r_{sphere}=const" lines converge.

**Looking at the black hole and world lines**

Of course, the remaining part of the picture contains the actual "black hole" that is so exciting. But let us start at the beginning, before the animal was born.

We see a world line of a point on the star surface: see the yellow line. The star has burned its fuel whose pressure kept it from the gravitational collapse. At some moment, the radius of the star drops below the Schwarzschild radius - that's where the yellow and orange lines intersect. At this moment, the star is pretty much doomed. Right now I am not sure whether one can generally prove that the event horizon becomes inevitable at this point but I am sure that it is going to be there in a moment according to the picture above which is surely realistic.

You should notice that after (="above" in the picture) the intersection of the yellow and orange lines, the star surface keeps on shrinking. At some moment, a point on the star surface - or the beer can - crosses the diagonal green line, the event horizon. The event horizon is the boundary between the (red) black hole interior and the rest of the spacetime.

This boundary is inevitably null i.e. light-like as you can easily see from the definition of the (red) black hole interior. It is the collection of points from which you can never get to the infinite future (scri plus) by following a time-like (mostly vertical, up) trajectory.

As you can see, there only exist two types of (timelike) world lines that are qualitatively different: the yellow world line, describing the star surface or the beer can, and the orange line, describing all observers such as Julio and Paolo who manage to escape from the gravitational grip of the black hole and survive at infinity.

If you want to stay outside the black hole but get really close to the horizon, your world line on the diagram will approach the green line, the event horizon, especially in the upper part of the green line where most of the spacetime volume (of the long black hole life, before it evaporates) is concentrated: the scale factor goes to infinity in all corners of the picture, with a possible exception of the left end of the singularity (I don't want to think too hard now), and in the whole "scri minus" and "scri plus".

On the other hand, if you lose your fight, like the beer can did, you will qualitatively follow the yellow line. When you cross the green event horizon, you don't even notice. But you should expect the last portion of your life inside the (red) interior to be finite, according to your proper time (clock).

You will experience increasing tidal forces (the gradient of the acceleration, using the old language) and right before the singularity (or earlier), you will be crushed into pieces. The Schwarzschild singularity is horizontal i.e. spacelike: that's because the timelike and spacelike directions are kind of interchanged inside the black hole because the factor (1-2GMr/c^2) flips the sign at the horizon. If you use the Schwarzschild coordinates, the line r=0 with t arbitrary is a spacelike, not timelike direction.

Once you're destroyed at the singularity, you die. Incidentally, the information about you must be secretly encoded and it must "superluminally tunnel" outside the black hole so that it is imprinted into the Hawking radiation at the "infinite future" because the information is not allowed to be lost at the singularity. But I don't want to discuss quantum gravity in this rudimentary posting.

The picture makes it clear that there is nothing surprising about the fact that the yellow and green lines intersect (much like the yellow and orange lines): a star can collapse so that all of its atoms are doomed to be destroyed at the singularity. They're surely doomed once they cross the horizon.

**The orange observer**

But you should really be interested in the orange observers who manage to stay outside the black hole. Because the black hole attracts them, they must accelerate away from the black hole. A spaceship would need powerful jets to escape the strong gravitational field, especially if it got really close to the event horizon.

In the Schwarzschild coordinates where the black hole looks static, you may be tempted to think that sitting at "r=const" is a natural state of affairs. However, there's nothing natural about it: "r=const" is not a geodesic: a "coordinate=const" line is rarely a geodesic in a curved spacetime. You need a nonzero acceleration to stay there. A spaceship would have to emit a lot of burned fuel to keep this position which would mean that its mass/energy would have to exponentially drop as a function of the proper time of the spaceship. But if you allow this "diet", a spaceship can sit there, of course.

If you want the spaceship to sit at "r=const" in Schwarzschild coordinates where "const" is just a little bit higher than the Schwarzschild radius, "r=2GM/c^2", you actually need a nearly infinite acceleration. You can see that the orange line approaches the green (null) line going in the Northwest direction as you approach the future (the upper part of the diagram).

Once the orange observer realizes that he must accelerate away from the black hole to save his life and freedom, his trajectory inevitably looks like the timelike hyperbola, "x^2 - c^2 t^2 = c^4/A^2" in the Minkowski space ("x" is positive).

This hyperbola describes an observer whose proper acceleration is constant: note that the hyperbola is given by a Lorentz-invariant equation so its form doesn't change if you switch to a different inertial frame, associated with the same observer after its velocity has accelerated a bit. The curvature radius of this hyperbola, "c^2/A", is proportional to the inverse acceleration (multiplied by the squared speed of light).

All the observers near the black hole horizon will be able to use the normal physics of special relativity because the curvature is very small there - if the black hole is large. General relativity always reduces to special relativity in spacetime regions that are much smaller than all the typical curvature radii. That's a statement that holds for any kind of the Riemannian geometry: in small enough local regions, it reduces to the Euclidean geometry, too. That's why the surface of the Earth is flat for most practical purposes. But these nearly flat "maps" can be glued into a globally curved picture.

If you introduce some natural, special-relativistic coordinates for such a small spacetime region near the horizon, in which the geometry is essentially "c^2 t^2 - x^2 - y^2 - z^2", you will see that the horizon itself is nothing else than a plane that is moving outwards by the speed of light, creating a null hypersurface in spacetime "x=ct" (flat Minkowski coordinates) and trying to eat everyone who hasn't escaped. This "x=ct" is the very same thing that you would describe as "r_{Schwarzschild} = 2GM/c^2" in Schwarzschild coordinates.

On the other hand, the massive observers who want to escape from the horizon have world lines analogous to "x^2 - c^2 t^2 = c^4/A^2". They rapidly approach the speed of light in these special relativistic coordinates that are locally valid. For a very small vicinity of these world lines, you may still introduce pretty good coordinates. But you should understand that the observers who are close to the horizon but who resist the gravity and stay outside are nothing like inertial observers. They're like heavily accelerating observers because their jets are on.

**Shrinking horizon**

There is one more point that may be confusing. I wrote that in the special relativistic coordinates describing a region near the black hole horizon, the horizon itself looks like a plane that moves outwards, i.e. away from the black hole center, trying to eat everyone who is not escaping.

On the other hand, you know that if the black hole evaporates, the radius of the event horizon should be slowly decreasing. It looks like the horizon is moving inwards. Is there a contradiction? Of course, there is no contradiction here.

In flat Euclidean geometry, the same coordinate "r" describes the proper distance from the origin as well as the circumference of the maximal circles on the sphere (divided by 2 pi) and other things. But in general relativity, there are infinitely many things that can be called "r" - there exists a freedom to choose coordinates - and the different types of "r" from the case of flat geometry no longer coincide.

For example, the circumference of the Earth's equator is not equal to pi times the distance (along the Earth's surface) of its antipodal points. Pi must be replaced by 2 here, in fact.

So when you look at the green event horizon and you follow it to the future - from the bottom up - you see that it is moving to the right side, which is analogous to an "increasing radial coordinate". However, there is a two-sphere sitting at each point of this green line and the radius of the two sphere is actually decreasing as you move towards the future. There's no paradox here: the proper radius of the hidden sphere is an independent function from the coordinates on our Penrose diagram. The only way to "prove" that their time derivatives should have the same sign would be to incorrectly assume that the geometry is flat: well, it's not.

**Summary**

The causal structure of a black hole may look paradoxical. The main reason is that an orange observer who succeeds in keeping himself outside the black hole had to be accelerating for a pretty long time which makes his world line "highly boosted" relatively to "natural" coordinates of observers who didn't have to accelerate for such long periods of time (e.g. because they either kept themselves very far from the black hole or they gave up and decided to fall inside).

The other point is that physics in small enough spacetime regions - except for the regions near the singularity - obeys the laws of special relativity pretty accurately. That's true even for the regions near the black hole event horizon. If you choose coordinates in which the local geometry looks like the Minkowski geometry in special relativity, the horizon becomes a plane moving outwards by the speed of light, trying to eat everyone. The observers who escape from it look like the constant-acceleration, hyperbolic world lines. And the black hole exterior looks like the Rindler wedge, i.e. the region with "x" and "x^2 - c^2 t^2" positive in the Minkowski space (with "y,z" arbitrary).

And the last thing you must realize is that in a curved spacetime geometry, the letters like "r" or "t" can mean many things. Different types of "radial coordinates" that agreed in the flat space differ once the geometry is curved, as we explained. But analogously, there are also many different meanings of "t": you can choose many types of the time coordinates. Even if you tell me that someone's proper time should be used and you tell me where the observer is located (i.e. a distance from the event horizon), it doesn't determine how quickly his time is going relatively to nearby objects in motion.

Just like in special relativity, you also need to know his velocity - the tangent direction of his world line. That's nothing new: special relativity must still be a limit of general relativity in small enough regions. The observers who live "right above" the event horizon see everything in a short proper time. But that's just because they move - morally almost by the speed of light - relatively to other observers. Different observers, such as the infalling beer can, view time very differently. After a very short time, they cross the horizon. Time is relative, just like it has been since 1905: observers at different velocities experience it differently.

I will probably not have time to check the text above and fix the mistake, at least not for half a day. So the text above is likely to be imperfect.

*Press F11 for full screen, to get more vertical space: another F11 gets you back. Use the arrow keys, Page Up/Down keys, the mouse button, or the mouse wheel to navigate through the box above, after you click inside it.*

## snail feedback (2) :

Dear Lubos,

I wonder if i can use the above image (bh diagram) for my personal perpose, which is not commercial at all. Please let me know!

Sure, no strings (or hair) attached.

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